4. The vertical parabolic curve has its P.I. at station 14 + 750 with elevation of 76.30m. The grade of the back tangent is 3.4% and the forward tangent is -4.8%. If the length of the curve is 300, a. Compute the location of the vertical curve turning point form P.I. (highest point). b. Compute the elevation of the vertical curve turning point. c. Compute the stationing of the vertical curve turning point.

Q: 23.15 A +4% grade intersects a -3% grade at station 3+ 000m at elevation 566.98 m. Compute the…

A: PVCSTATION=PVISTA-L2PVCsta=3000-6002=2+700PVCELV=PVIELV-g1L2PVCelv=566.98-0.04×6002PVCelv=554.98m

Q: Problem 4 A vertical summit parabolic has its P.I. at station 12 + 750 with elevation of 77.30 m.…

A: Given Data:Station of PV1=12+750Elevation Pv1=77.30mForward Tangent g2=-3.8%back tangent…

Q: Vertical Curve 4.) The same section of road is also on a vertical curve. Where L= 200 G1 = 2% and G2…

A:

Q: 23.1 A 1,500-ft vertical curve is to be used for joining a +3.0% grade line and a-4.0% grade line.…

A: The length of curve, L= 1500 ftThe grade of upgrade, g1 = +3%The grade of downgrade, g2 = -4%The…

Q: 2. Vertical highway curve: Given PVI at 19+00, L = 500 ft, g1 = +2.5%, g2=-1.0%, and PVI elevation…

A:

Q: Problem 2 A symmetrical vertical summit curve has tangents of +5% and -3%. The allowable rate of…

A:

Q: 2. Vertical Curve Given the following vertical curve data, compute the elevations of the curve,…

A: Given: Given the following vertical curve data             g1= -1%, g2 = 1.5%, L=100m, PVI at 2+200,…

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